If you add and subtract x^2 from the numerator you can split up the fraction in to 1 + (x^5 - x^2) / (x^2+x+1) = 1 + x^2 * (x^3 - 1) / (x^2+x+1) = 1 + x^2 * (x - 1)(x^2 + x + 1)/ (x^2+x+1) = 1+ x^3 - x^2 Final result x^3 - x^2 + 1 There's also a way of doing this by factoring the numerator but it's gonna be a bit of brute force and systems and a whole lot of work, and maybe some guessing. It's a nice problem that rewards you for good observation making.
@udic01
Ай бұрын
Just add (and subtract) x^4+x^3+x^2...
@archnof0
Ай бұрын
Yeah either that or synthetic division according to your mood
@dhruvchaudhary5379
Ай бұрын
Add and subtract x^2 in numerator to get x^2(x-1) +1
@CalculusIsFun1
Ай бұрын
I just did long division. looks something like this. the x^2 + x + 1 goes into it x^3 number of times leaving remainder of -x^4 - x^3 + x + 1 which then goes into it -x^2 number of times leaving remainder x^2 + x + 1 which goes into it 1 time. Final answer is x^3 - x^2 + 1
@hevado01
Ай бұрын
My method too, appeared quite simple
@terryendicott2939
Ай бұрын
Multiply both top and bottom by (x-1). The top is x^6-1 and the bottom is x^3-1. Factori both into the complex roots of one and watch the carnage of cancellation.
@yuryp6975
Ай бұрын
The numerator will not be X^6-1
@terryendicott2939
Ай бұрын
@@yuryp6975 Well I feel stupid. Thanks--- I was thinking x^5+x^4+x^3+x^2+x+1 just missed a few terms.
@robertveith6383
Ай бұрын
@@yuryp6975 Your upper case letters represent different variables from the user's lower case letters.
@matthewfeig5624
Ай бұрын
The numerator will have x^6-1 and two extra terms, but this approach still works well. Num = x^6 - 1 - x^5 + x^2 = (x^3+1)(x^3-1) - x^2 (x^3-1) = (x^3+1-x^2)(x^3-1)
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